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Reinforced concrete beam with web opening 2

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TheDW

Structural
Feb 17, 2016
146
I'm using this paper to check our detail for a beam with very high shear demand right at a penetration:
And I'm stuck on equation 3:

Beam_type_failure_aonleg.png


Am I misreading it that the Vc contribution goes from 2*sqrt(f'c)*b*d to 1/6*sqrt(f'c)*b*d_reduced? I would think that I would just use 2sqrt(f'c)*b*d_reduced.

I've got a 6ksi 48"w x 36"d (d=31.76) beam with a 4" penetration centered and this takes my Vc from 206k to 17k.

The paragraph above the equation mentions the simplified approach from ACI 1995 being the source of the equation, but it is nowhere to be found in the shear/torsion chapter - I'm sure I'm missing something simple, can anyone help me out?
 
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Assuming that the units for the equation you're showing is in psi, that does seem really low. It would be really high for it in ksi, though.

Edited post - in my original post, I was thinking of the AASHTO Bridge design equations, which are in ksi, so I got all confused.
 
You'd have to purchase the 1998 Mansur paper to understand it - I couldn't find any "free" reference out there that talked about the 1/6 value.
You can find the paper online with a search but you have to pay for the pdf of it.

It does seem like a large reduction in capacity but with the hole breaking up the remaining Vc into smaller segments I guess I can understand it.

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I believe the 1/6 coefficient is for SI units. The SI equivalent of that coefficient in ACI 318-11 is 0.17, which is 1/6 rounded off to two decimal places.
 
Thanks Hokie. Now it all makes sense - it's all the same equation in different unit systems (including the AASHTO equation I'm familiar with, which uses a 0.0316 coefficient for f'c in ksi).
 
Nice, its always the units...thanks Hokie
 
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